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Unformatted text preview: Flores, Michael – Homework 1 – Due: Jan 29 2004, 4:00 am – Inst: Sonia Paban 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. This set of homework covers electrostatic problems involving discrete point charges. Problems involving continuous charge distri bution will be covered in the next homework. For electrostatic problems, unless otherwise stated for Coulomb constant, please use : k e = 8 . 9875 × 10 9 Nm 2 /C 2 001 (part 1 of 1) 10 points A neutral metal ball is suspended by a string. A positively charged insulating rod is placed near the ball, which is observed to be at tracted to the rod. This is because 1. the ball becomes negatively charged by induction. 2. the ball becomes positively charged by induction. 3. the string is not a perfect insulator. 4. the number of electrons in the ball is greater than in the rod. 5. there is a rearrangement of the electrons in the ball. correct Explanation: Electrons in the metal ball will be attracted toward the charged insulating rod. 002 (part 1 of 1) 10 points Given: k e = 8 . 9875 × 10 9 N m 2 / C 2 . A particle with charge 4 μ C is located on the xaxis at the point 10 cm , and a second particle with charge 7 μ C is placed on the xaxis at 4 cm . 2 4 6 8 10 2 4 6 8 10 4 μ C 7 μ C 3 μ C x → (cm) What is the magnitude of the total elec trostatic force on a third particle with charge 3 μ C placed on the xaxis at 2 cm ? Correct answer: 464 . 354 N. Explanation: Basic Concept: Coulomb’s law (in vector form) for the electric force exerted by a charge q 1 on a second charge q 3 , written ~ F 13 is ~ F 13 = k e q 1 q 3 r 2 ˆ r 13 , where ˆ r 13 is a unit vector directed from q 1 to q 3 ; i.e. , ~r 13 = ~r 3 ~r 1 . Let : q 1 = 4 μ C = 4 × 10 6 C , q 2 = 7 μ C = 7 × 10 6 C , q 3 = 3 μ C = 3 × 10 6 C , x 1 = 10 cm = 0 . 1 m , x 2 = 4 cm = . 04 m , x 3 = 2 cm = . 02 m , so x 13 = x 3 x 1 = ( 2 cm) (10 cm) = . 12 m x 23 = x 3 x 2 = ( 2 cm) ( 4 cm) = 0 . 02 m ˆ x 13 = ( x 3 x 1 ) p ( x 3 x 1 ) 2 = ˆ ı ˆ x 23 = ( x 3 x 2 ) p ( x 3 x 2 ) 2 = +ˆ ı Since the forces are collinear, the force on the third particle is the algebraic sum of the forces between the first and third and the second and third particles. ~ F = ~ F 13 + ~ F 23 = k e • q 1 r 2 13 ˆ r 13 + q 2 r 2 23 ˆ r 23 ‚ q 3 = (8 . 9875 × 10 9 N m 2 / C 2 ) × • (4 × 10 6 C) ( . 12 m) 2 ( ˆ ı ) + (7 × 10 6 C) (0 . 02 m) 2 (+ˆ ı ) ‚ × (3 × 10 6 C) = ( 7 . 48958 N) + (471 . 844 N) = 464 . 354 N k ~ F k = 464 . 354 N . Flores, Michael – Homework 1 – Due: Jan 29 2004, 4:00 am – Inst: Sonia Paban 2 003 (part 1 of 1) 10 points Suppose that 3 . 5 g of hydrogen, H 2 , is sep arated into electrons and protons, and that the protons are placed at the Earth’s North Pole and the electrons are placed at the South Pole. The radius of the Earth is 6Pole....
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This note was uploaded on 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.
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